In right triangle \( \mathrm{ABC}, \mathrm{m} \angle A=90^{\circ}, \cos B=4 x-0.2 \), and \( \sin C=7 x-0.14 \). Solve for \( x \).

In triangle \( \mathrm{ABC} \) where \( \angle A = 90^\circ \), we have \( \angle B + \angle C = 90^\circ \) (since the angles in a triangle sum up to \( 180^\circ \)). This relationship tells us that \( \sin C = \cos B \). Given \( \cos B = 4x - 0.2 \) and \( \sin C = 7x - 0.14 \), we can set these two expressions equal since \( \sin C = \cos B \): \[ 4x - 0.2 = 7x - 0.14. \] Now, let's solve for \( x \): 1. Rearranging the equation gives: \[ 4x - 7x = -0.14 + 0.2, \] which simplifies to: \[ -3x = 0.06. \] 2. Dividing both sides by -3 results in: \[ x = -\frac{0.06}{3} = -0.02. \] Thus, the solution for \( x \) is: \[ \boxed{-0.02}. \]